Kinetics of the Formation of HBr

Kinetics of the Photochemical Reaction Between Hydrogen and Bromine

The photochemical combination of moist H₂ and Br₂ vapor in visible light(< 510nm) is a chain reaction and is occurs at 433-491 k.
H₂ + Br₂ ---h𝜈→ 2HBr
Possible mechanism of this photochemical reaction-
Chain Initiation-
1. Br₂ + h𝜈 ---k₁→ 2Br
Chain Propagation-
ii. Br + H₂ ---k₂→ HBr + H
iii. H + Br₂ ---k₃→ HBr + Br
Chain Inhibition-
iv. H + HBr ---k₄→ H₂ + Br
Chain Termination-
v. Br + Br ---k₅→ Br₂
where, k₁, k₂, k₃, k₄ and k₅ are rate constants.

Since HBr is formed in steps 'ii' and step 'iii' and disappear in step 'iv', hence, the net rate of formation of HBr-
d[HBr]/dt = k₂[H₂][Br] + k₃[H][Br₂] − k₄[H][HBr]  ---Eq-1
The H atom are formed in step'ii' and disappear in steps 'iii' and 'iv', hence-
d[H]/dt = k₂[Br][H₂] − k₃[H][Br₂] − k₄[H][HBr]  ---Eq-2
Applying steady state approximation, we get-
0 = k₂[Br][H₂] − k₃[H][Br₂] − k₄[H][HBr]
or, k₂[Br][H₂] = k₃[H][Br₂] + k₄[H][HBr]

The Br atoms are formed in steps 'i', 'iii' and 'iv' and disappear in step 'ii' and 'v', hence-
d[Br]/dt = k₁I_abs − k₂[Br][H₂] + k₃[H][Br₂] + k₄[H][HBr] − k₅[Br]²
Applying steady state approximation, we get-
k₁I_abs + k₃[H][Br₂] + k₄[H][HBr] = k₂[H₂][Br] + k₅[Br]²  ---Eq-3

Subtracting equation-2 from equation-3 we get-
k₁I_abs = k₅[Br]²
or, [Br] = (k₁I_abs/k₅)^1/2
Putting the value of [Br] in equation-2 we get-
k₂(k₁I_abs)^1/2[H₂] = k₃[H][Br₂] + k₄[H][HBr]


This equation agree with the experimental value. Therefore, the rate of the reaction varies with the square root of the intensity of light(I_abs).

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Q. Show that the rate of reaction varies directly to the square root of the intensity of radiation absorbed.

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